Hadamard Gap Theorem | Real Analysis

Exponents doubling like 1, 2, 4, 8 leave bigger and bigger holes. The holes block continuation to every point of the circle. This happens even though the series converges nicely inside. The gap-free comparison series stays calm.

For example, imagine a 🪜 staircase whose steps spread by 1, 2, 4, 8. The first steps are easy. Soon one step is too far, and the wall's end is out of reach.

Examples & Exercises

Prove by the Cauchy-Hadamard root test that $f(z) = \sum_{k=0}^{\infty} z^{2^k}$ has radius of convergence exactly 1.

f(z) = \sum_{n=0}^{\infty} a_n z^n, \qquad a_n = 1\ (n = 2^k),\; a_n = 0\ \text{otherwise}

\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n \rvert^{1/n} = 1 \;\Rightarrow\; R = 1

N = 5

: inside the disc,

S_5(0.5) \approx 0.81642150902

converges fast — consistent with the radius-1 proof below.

Show proof

Write $f(z)$ as an ordinary power series $\sum_{n=0}^{\infty} a_n z^n$ : $a_n = 1$ when $n = 2^k$ and $a_n = 0$ otherwise. The Cauchy-Hadamard formula gives $\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n \rvert^{1/n}$ (with the conventions $1/0 = \infty$ , $1/\infty = 0$ ).
Since $\lvert a_n \rvert^{1/n} \le 1$ for every $n \ge 1$ (at $n = 0$ , $1/n$ is undefined, so $n = 0$ is excluded from the limsup argument), the limsup is at most 1.
But along the subsequence $n = 2^k$ ( $k \ge 1$ ) we have $\lvert a_{2^k} \rvert^{1/2^k} = 1^{1/2^k} = 1$ , so the limsup is at least 1.
Hence the limsup equals 1 and $R = 1$ — exactly 1.
So the series converges absolutely for $\lvert z \rvert < 1$ , and diverges for $\lvert z \rvert > 1$ since the general term $z^{2^k}$ does not tend to 0 ( $\lvert z \rvert^{2^k} \to \infty$ ). The boundary $\lvert z \rvert = 1$ is where the test goes silent — exactly where the gap theorem speaks.

Prove by the Weierstrass M-test that $T(z) = \sum_{k=0}^{\infty} \frac{z^k}{(k+1)^2}$ converges uniformly on the closed unit disc $\lvert z \rvert \le 1$ , so its sum extends continuously to the boundary.

Unlike the gap series, this sum is continuous on the closed disc, so $\lvert z \rvert = 1$ is not a natural boundary for the continuous extension (claiming no analytic singularity would go beyond this proof).

\left\lvert \frac{z^k}{(k+1)^2} \right\rvert \le \frac{1}{(k+1)^2} = M_k, \quad \sum_{k=0}^{\infty} M_k = \frac{\pi^2}{6} < \infty

\max_\theta \lvert T_{200}(0.99\,e^{i\theta}) \rvert \approx 1.6045 < 1.65 < \frac{\pi^2}{6}

N = 12

: the dashed gray

T_{12}

stays calm at

\max_\theta \lvert T_{12} \rvert \approx 1.5551

— without gaps there is no blow-up at the boundary.

Show proof

On $\lvert z \rvert \le 1$ , $\left\lvert z^k/(k+1)^2 \right\rvert \le 1/(k+1)^2 =: M_k$ , and $\sum M_k = \sum_{k=1}^{\infty} 1/k^2 = \pi^2/6 < \infty$ (the Basel problem). By the Weierstrass M-test, $\sum z^k/(k+1)^2$ converges uniformly on the closed disc $\lvert z \rvert \le 1$ .
Each partial sum is a polynomial, hence continuous, and a uniform limit of continuous functions is continuous — so the sum $T(z)$ is continuous on the whole closed disc.
That is, $T$ extends continuously to the boundary $\lvert z \rvert = 1$ : every boundary point is a point of continuity for the continuous extension.
That is exactly what this proof shows: continuity alone does not imply analytic extension or absence of an analytic singularity (for example $T(z) = \mathrm{Li}_2(z)/z$ extends continuously, yet $z = 1$ is a singularity of $\mathrm{Li}_2$ ).
$T(z)$ and $\sum z^{2^k}$ both have radius exactly 1 yet opposite fates — the gaps make the difference. The measurements agree: $\max_\theta \lvert T_{200}(0.99\,e^{i\theta}) \rvert \approx 1.6045 < \pi^2/6 \approx 1.6449$ , with no hint of blow-up near the boundary.

Sketch the key idea of the Hadamard gap theorem in 2–3 steps: assuming a holomorphic extension past the boundary, explain the mechanism by which the gaps force a contradiction.

State explicitly in your answer that this is a sketch of the idea, not the complete proof.

s_k(z) = \sum_{j=0}^{k} z^{2^j}, \qquad s_{k+1}(z) - s_k(z) = z^{2^{k+1}}

\lvert z_0 \rvert > 1 \;\Rightarrow\; \lvert z_0 \rvert^{2^{k+1}} \to \infty

N = 12

: 106 spikes crowd the boundary — below is the proof idea behind what this picture suggests (a sketch, not the complete proof).

Show hint

Start with contradiction: suppose $f(z) = \sum_{k\ge 0} z^{2^k}$ extended holomorphically past some boundary point $\zeta$ ( $\lvert \zeta \rvert = 1$ ). Then there is a small disc $D$ around $\zeta$ on which a holomorphic function agrees with $f$ inside $\lvert z \rvert < 1$ — so $D$ pokes outside the unit circle.
[IDEA — the step not proved here] Gap-forced overconvergence: the partial sums $s_k(z) = \sum_{j=0}^{k} z^{2^j}$ converge to $f$ on $\lvert z \rvert < 1$ , hence to the extension on $D \cap \{\lvert z \rvert < 1\}$ .
Each new partial sum adds just one term yet leaps twice as far in degree ( $2^{k+1} = 2\cdot 2^k$ ) — with no intermediate terms to interfere, a classical overconvergence lemma forces the sequence $s_k$ to keep converging uniformly on all of $D$ , including points with $\lvert z \rvert > 1$ .
That lemma is the technical heart; it is stated here, not proved.
But at any point with $\lvert z_0 \rvert > 1$ , $s_{k+1}(z_0) - s_k(z_0) = z_0^{2^{k+1}}$ and $\lvert z_0^{2^{k+1}} \rvert = \lvert z_0 \rvert^{2^{k+1}} \to \infty$ , so $s_k(z_0)$ cannot converge — a contradiction.
Hence no holomorphic extension past any boundary point exists, and every point of the unit circle is singular.
To stress again: this is the idea (the skeleton) of the proof, not the complete proof.